Answer
$\dfrac{2\sin x\cos x}{(\sin x+\cos x)^{2}-1}=1$
Work Step by Step
$\dfrac{2\sin x\cos x}{(\sin x+\cos x)^{2}-1}=1$
On the left side of the equations, evaluate $(\sin x+\cos x)^{2}$:
$\dfrac{2\sin x\cos x}{\sin^{2}x+2\sin x\cos x+\cos^{2}x-1}=1$
Since $\sin^{2}x+\cos^{2}x=1$, the expression becomes:
$\dfrac{2\sin x\cos x}{1-1+2\sin x\cos x}=1$
Simplify and the identity is proved:
$\dfrac{2\sin x\cos x}{2\sin x\cos x}=1$
$1=1$
